The user wants me to solve a chemistry problem involving a stoichiometric calculation.

Step-by-Step Solution:

1. Write the balanced chemical equation: The reaction between diborane ($B_2H_6$) and an alcohol, such as methanol ($CH_3OH$), results in the formation of a trialkyl borate and hydrogen gas. This reaction proceeds via the cleavage of the B-H bonds in diborane. The balanced chemical equation for the reaction is: $B_2H_6 + 6 CH_3OH \rightarrow 2 B(OCH_3)_3 + 6 H_2$

2. Identify the reactants and products:

   • Reactants: Diborane ($B_2H_6$) and Methanol ($CH_3OH$).
   • Products: Trimethyl borate ($B(OCH_3)_3$) and Hydrogen gas ($H_2$). The question asks for the number of moles of the boron-containing product, which is trimethyl borate, $B(OCH_3)_3$.

3. Analyze the stoichiometry of the reaction: From the balanced chemical equation, we can establish the molar ratio between the reactant diborane ($B_2H_6$) and the boron-containing product trimethyl borate ($B(OCH_3)_3$). The ratio is: $1 \text{ mole of } B_2H_6 : 2 \text{ moles of } B(OCH_3)_3$

4. Calculate the number of moles of the product: The problem states that 3 moles of $B_2H_6$ are completely reacted. Using the stoichiometric ratio from step 3, we can calculate the moles of $B(OCH_3)_3$ produced. $\text{Moles of } B(OCH_3)_3 = (\text{Moles of } B_2H_6) \times \frac{2 \text{ moles } B(OCH_3)_3}{1 \text{ mole } B_2H_6}$ Substituting the given value: $\text{Moles of } B(OCH_3)_3 = 3 \text{ moles} \times 2$ $\text{Moles of } B(OCH_3)_3 = 6 \text{ moles}$

5. Final Answer: Therefore, the complete reaction of 3 moles of $B_2H_6$ with methanol produces 6 moles of the boron-containing product, trimethyl borate.